Multi-Level Approaches to the Solution of Problems
How a student answers a question is more important than the answer given by the student. For example, the student may have randomly guessed, the student may have used a rote and unimaginative method for solution, or the student may have used a very creative method. It seems that one should judge the student by the “way” he or she answers the question and not just by the answer to the question. Example: Question: Without using a calculator, which is greater: 355 x 356 x 354 or 354 x 357? Case 1: 'Rote Memory Approach '(a completely mechanical approach not realizing the fact that there may be a faster method that takes into account patterns or connections of the numbers in the question): The student multiplies 355 x 356 gets 126,380, and then multiplies 354 x 357 and gets 126 x 378. Case 2: 'Observer's Rot Approach '(an approach that makes use of a mathematical strategy that can be memorized and tried for various problems): The student does the following: Divide both quantities by 354: He or she then gets 355 x 356/354 compared with 354 x 357/354. He or she then divides these quantities by 356 and then gets 355/354 compared with 357/356. Now he or she realizes that 355/354 = 1 and 1/354; 357/356 = 1 and 1/356. He or she then reasons that since the left side 1 and 1/354 is greater than the right side, 1 and 1/356, the left side of the original quantities, 355 x 356 is greater than the right side of the original quantities, 354 x 357. Case 3: 'The Pattern Seeker's Mind '(most mathematically creative method-- an approach in which the student looks for a pattern or sequence in the numbers and then is astute enough to represent the pattern or sequence in more general algebraic language to see the pattern or sequence more clearly): Look for a pattern. Represent 355 x 356 and 354 x 357 by symbols. Let x ''= 354. Then 355 = ''x + 1, 356 ='' x'' + 2, 357 = x ''+ 3. So 355 x 356 = (''x + 1) (x ''+ 2) and 354 x 357 = ''x(x + ''3). Multiplying the factors we get 355 x 356 = (''x times x'') + 3''x + 2 and 354 x 357 = (x ''times ''x) + 3''x'' + 2 minus (x'' times ''x) minus 3''x'', which is just 2. So 355 x 356 is greater than 354 x 357 by 2. Note: You could have also represented 355 by x''. Then 356 = ''x + 1;354 = x – 1;357 = x'' + 2. We could then get 355 x 356 = (''x) (x ''+ 1) and 354 x 357 = (''x – 1) (x'' + 2). Then we would use the method above to compare the quantities. *OR – You could have written 354 as ''a ''and 357 as ''b. ''Then 355 = ''a ''+ 1 and 356 = ''b – 1. So 355 x 356 = (a ''+ 1) (''b – 1) and 354 x 357. Let's see what (355 x 356) – (354 x 357) is. This is the same as (a ''+ 1) (''b – 1) – ab, which is (ab ''+ ''b – a – 1) –'' ab'', which is in turn b – a – ''1. Since ''b – a – ''1 = 357 – 354 – 1 = 2, the quantity 355 x 356 – 354 x 357 = 2, so 355 x 356 is greater than 354 x 357 by 2. ''Case 4: 'The Astute Observer's Approach '(the simplest approach-- an approach that attempts to figure out a connection between the numbers and uses that connection to figure out the solution): 355 x 356 = (354 + 1) x 356 =(354 x 356) + 356 and 354 x 357 = 354 x (356 + 1) = (354 x 356) + 354 One can see that the difference is just 2. Case 5: 'The Observer's Common Relation Approach '(this is the approach that people use when they want to connect two items to a third to see how the two items are related): 355 x 356 is greater than 354 x 356 by 356. 354 x 357 is greater than 354 x 356 by 354. So this means that 355 x 356 is greater than 354 x 357. Case 6: 'Scientific, Creative, and Observational Generalization Method '(a highly creative method and the most scientific method, as it spots a critical and curious aspect of the sums being equal and provides for a generalization to other problems of that nature): Represent 354 = a'', 357 = ''b, 355 = c'', and 356 = ''d We have now that (1) a + b ''= ''c + d #|''b – a''| > |''d – c''| We want to prove ab ''> ''dc Proof: Square inequality (2): (b – a'')2 > (''d – c)''2 Therefore: (3) ''b''2 – 2''ab + a2 ''> d''2'' – 2''dc + c''2 Multiply (3) by (– 1) and this reverses the inequality sign: *(''b''2 – 2''ab ''+ ''a2'') < – (''d''2 – 2''dc – c''2) or #– ''b''2 + 2''ab – a''2 < – ''d''2 + 2''dc –c''2 Now square (1): (''a ''+ ''b) = (c ''+ ''d) and we get: #''a''2 + 2''ab + ''b''2 = ''c''2 + 2''dc ''+ ''d2'' Add inequality (4) to inequality (5) and we get 4''ab ''< 4''dc Divide by 4 and we get: ab > dc The generalization is that for any positive numbers a, b, c, d ''when |''b – a| > |d – c''| and ''a ''+ ''b ''= ''c ''+ ''d, ''then ''ab < dc. This also generalizes in a geometrical setting where for two rectangles whose perimeters are the same (2''a + 2''b ''= 2''c ''+ 2''d), the rectangle whose absolute difference in sides |''d – c| ''is least has the greatest area. Category:Strategies Category:Introduction Category:SAT Introduction